Question

Hard water often contains dissolved Ca2 and Mg2 ions. One way to soften water is to add phosphates. The phosphate ion forms insoluble precipitates with calcium and magnesium ions, removing them from solution. Suppose that a solution is 0.052 M in calcium chloride and 0.093 M in magnesium nitrate. What mass of sodium phosphate would have to be added to 1.6 L of this solution to completely eliminate the hard water ions

Answer 1

Answer:

mass of sodium phosphate required = 25.37 g

Explanation:

Equations of reactions:

3CaCl₂ + 2Na₃PO₄ ---> Ca₃(PO₄)₂ + 6NaCl

3Mg(NO₃)₂ + 2Na₃PO₄ ---> Mg₃(PO₄)₂ + 6NaNO₃

Number of moles of the ions = molarity x volume

Ca²⁺ : 0.052 M x 1.6 L = 0.0832 moles

Mg²⁺ : 0.093 x 1.6 L = 0.1488 moles

From the equations of reaction, number of moles of sodium phosphate required to react with each ion is given as;

For calcium: 2/3 x 0.0832 = 0.0555 moles

For magnesium: 2/3 x 0.1488 = 0.0992 moles

Total number of moles of sodium phosphate required = 0.0555 + 0.0992 = 0.1547 moles

Mass of sodium phosphate = number of moles x molar mass

molar mass of sodium phosphate = 164 g/mol

Mass of sodium phosphate = 0.1547 moles x 164 g/mol = 25.37 g

Therefore, mass of sodium phosphate required = 25.37 g